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Axis of rotation, plane of reflection

  1. Oct 4, 2011 #1
    For a 3x3 orthogonal matrix with determinant= -1 (which means rotation followed by simple reflection), is the axis of rotation the same as the plane of reflection ?

    My reasoning is follows (see attachment)

    Say you have two vectors with the same angle size (which i call A), same x-values, but one of the z-values (height in this case), is the negative of the other (n.b. y-value is zero in both vectors )

    Vector 1 rotates around the x-axis by 2A to get to the same spot as vector 2. Because the angle size is the same and because the z-component of vector 2 is the negative of the z-component of vector 1, we get a reflection ?

    Since the "reflection" happens about the x-axis, this is why the plane of reflection is the same as the axis of rotation in the case of a 3x3 orthogonal matrix having determinant 1 ?

    Attached Files:

  2. jcsd
  3. Oct 6, 2011 #2
    if [itex]A\in \textrm{O}(3,\mathbb{R})[/itex] and [itex]\det(A)=-1[/itex], there exists a [itex]U\in \textrm{SU}(3)[/itex] such that

    UAU^{\dagger} = \left(\begin{array}{ccc}
    -1 & 0 & 0 \\
    0 & e^{i\theta} & 0 \\
    0 & 0 & e^{-i\theta} \\

    with some [itex]\theta\in\mathbb{R}[/itex].

    Then there exists a [itex]V\in \textrm{SU}(2)[/itex] such that

    e^{i\theta} & 0 \\
    0 & e^{-i\theta} \\
    = \left(\begin{array}{cc}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta) \\

    So if you define

    W = \left(\begin{array}{cc}
    1 & 0 \\
    0 & V \\

    then [itex]WAW^{\dagger}[/itex] will be of such form that reflection and rotation are clearly carried out with respect to the same axis. Only problem is that [itex]W[/itex] doesn't necessarily have only real entries. How to prove that [itex]W[/itex] is necessarily proportional to a real matrix?
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